title: “chapter4.Rmd” output: html_document author: Faranak Halali, 24.11.2019
Week 4: Clustering and classification
This week’s task deals with clustering and classification methods. Boston dataset will be used.
Loading and exploring data
library(MASS)
data("Boston")
str(Boston)## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...dim(Boston)## [1] 506 14Interpretation: Boston dataset contains 506 rows (observations) with 14 variables (columns). It is about housing values in suburbs of Boston, USA. All the variables are numeric. Example variables are: crim = per capite crime rates, indus = proportion of non-retail business acres per town, nox = nitrogen oxides concentration (parts per 10 million).
Graphical overview and summary of the data
summary(Boston)## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00pairs(Boston)
Interpretation: Summary of the variables shows the min and max values of each variable plus the quartiles (Q1, Q2 = median, Q3) and mean. For example for the variable crim: the mean value for the per capite crime rate is 3.61. The plot matrix shows the relationships between pairs of variables. For example, there is a low-steep positive relationship between nox and age.
Standardize the dataset
boston_scaled <- scale(Boston)
summary(boston_scaled)## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865class(boston_scaled)## [1] "matrix"boston_scaled <- as.data.frame(boston_scaled)Interpretation: Here, we standardized (z-normalized) the dataset because different variables in the dataset may have different dimensions and measured on different scales. Looking into the results of data normalization, all the variables have now a mean of zero.
Create a categorical variable of the crime rate
bins <- quantile(boston_scaled$crim)
bins## 0% 25% 50% 75% 100%
## -0.419366929 -0.410563278 -0.390280295 0.007389247 9.924109610crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high"))
table(crime)## crime
## low med_low med_high high
## 127 126 126 127Interpretation: Based on the its quantiles, the original continuous crim variables was converted to a 4-level categorical variable called crime. The categories of this new variable are: low, med_low, med_high, high.
Drop the old crime variable and add the new categorical crime variable
boston_scaled <- dplyr::select(boston_scaled, -crim)
boston_scaled <- data.frame(boston_scaled, crime)Divide the dataset to train and test sets
n <- nrow(boston_scaled)
ind <- sample(n, size = n * 0.8)
train <- boston_scaled[ind,]
test <- boston_scaled[-ind,]Save the correct classes from test data
correct_classes <- test$crimeRemove the crime variable from test data
test <- dplyr::select(test, -crime)Linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)
lda.fit## Call:
## lda(crime ~ ., data = train)
##
## Prior probabilities of groups:
## low med_low med_high high
## 0.2500000 0.2574257 0.2301980 0.2623762
##
## Group means:
## zn indus chas nox rm
## low 0.95572404 -0.8717686 -0.077423115 -0.8659269 0.4243107
## med_low -0.08320563 -0.2841228 -0.007331936 -0.5354109 -0.1371654
## med_high -0.38396622 0.1742905 0.193349455 0.4089740 0.2369176
## high -0.48724019 1.0170298 -0.012331882 1.0594552 -0.3723712
## age dis rad tax ptratio
## low -0.8673354 0.8359584 -0.6987343 -0.7469021 -0.40935287
## med_low -0.3178123 0.3271295 -0.5434660 -0.4529740 -0.09804605
## med_high 0.4338051 -0.3700372 -0.4101075 -0.2941221 -0.27200081
## high 0.8179427 -0.8481723 1.6390172 1.5146914 0.78181164
## black lstat medv
## low 0.3795874 -0.77254980 0.5165063
## med_low 0.3163503 -0.08680645 -0.0202947
## med_high 0.1445800 -0.04448735 0.2382108
## high -0.8182524 0.91972122 -0.7192353
##
## Coefficients of linear discriminants:
## LD1 LD2 LD3
## zn 0.11584078 0.72013904 -1.0322672
## indus 0.01601055 -0.23640390 0.1711228
## chas -0.09513173 0.01919263 0.1404721
## nox 0.44781528 -0.80406056 -1.2890937
## rm -0.10503721 -0.18839993 -0.1001859
## age 0.20836547 -0.38778595 -0.2733285
## dis -0.07954802 -0.40979808 0.1570075
## rad 3.44394872 1.04639786 -0.3080947
## tax 0.02133046 -0.19994256 0.8278850
## ptratio 0.12992525 0.02247724 -0.3680662
## black -0.13213314 0.02903654 0.1365207
## lstat 0.21135334 -0.19235218 0.5662801
## medv 0.20352002 -0.43355471 -0.1759290
##
## Proportion of trace:
## LD1 LD2 LD3
## 0.9571 0.0307 0.0122Interpretation: out of 404 observations (5060.80), 100 belong to low crime category, 99 to med_low, 105 to med_high, and 100 to high category. Next, we see the means of scaled variables in each of the 4 crime categories. Coefficients of linear discriminants show the linear combination of predictor variables that are used to form the LDA decision rule. for example, LD1 = 0.132zn + 0.039indus - 0.069chas,…… . Proportion of trace is the percentage separation in crime classes achieved by each discriminant function. In this case, LD1 has the highest separation percentage.
Plot the LDA results
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
classes <- as.numeric(train$crime)
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1)
Interpretation: I can say that the combination of ‘rad & zen & nox’ variables are the most influential separators of the crime categories.
Predict LDA
lda.pred <- predict(lda.fit, newdata = test)
table(correct = correct_classes, predicted = lda.pred$class)## predicted
## correct low med_low med_high high
## low 17 8 1 0
## med_low 2 18 2 0
## med_high 1 13 17 2
## high 0 0 0 21Interpretation: LDA estimates the probability of a new set of inputs belonging to every class. The output class is the one that has the highest probability. In the test data, 32 observations are in the low crime category and LDA predicts 16 of them to be in low and the other 16 to be in med_low crime category (50% correct prediction). Out of 24 observations in the med_low crime category the LDA model predicts 20 in med_low and 4 in low category (83% correct prediction). Out of 24 med_high observations, the model predicts 12 in med_low and 12 in med_high category (50% correct prediction). Of 21 high crime observations the model predicts all to belong to high category (100% correct prediction). So, the highest correct predictions belong to the 2 ends of the crime levels, i.e., low and high.
Reload and standardize the Boston dataset
library(MASS)
data("Boston")
boston_scaled <- scale(Boston)
summary(boston_scaled)## crim zn indus
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668
## Median :-0.390280 Median :-0.48724 Median :-0.2109
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202
## chas nox rm age
## Min. :-0.2723 Min. :-1.4644 Min. :-3.8764 Min. :-2.3331
## 1st Qu.:-0.2723 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366
## Median :-0.2723 Median :-0.1441 Median :-0.1084 Median : 0.3171
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.:-0.2723 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059
## Max. : 3.6648 Max. : 2.7296 Max. : 3.5515 Max. : 1.1164
## dis rad tax ptratio
## Min. :-1.2658 Min. :-0.9819 Min. :-1.3127 Min. :-2.7047
## 1st Qu.:-0.8049 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876
## Median :-0.2790 Median :-0.5225 Median :-0.4642 Median : 0.2746
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6617 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058
## Max. : 3.9566 Max. : 1.6596 Max. : 1.7964 Max. : 1.6372
## black lstat medv
## Min. :-3.9033 Min. :-1.5296 Min. :-1.9063
## 1st Qu.: 0.2049 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median : 0.3808 Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.4332 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 0.4406 Max. : 3.5453 Max. : 2.9865class(boston_scaled)## [1] "matrix"boston_scaled <- as.data.frame(boston_scaled)Calculate the Euclidian and Manhattan distances between the observations
dist_eu <- dist(boston_scaled)
summary(dist_eu)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4625 4.8241 4.9111 6.1863 14.3970dist_man <- dist(boston_scaled, method = 'manhattan')
summary(dist_man)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.2662 8.4832 12.6090 13.5488 17.7568 48.8618Interpretation: I calculated the similarity (distance) between the observations by two methods: Euclidian distances (EuD) and Manhattan distances (MD). EuD is the length of the line segment connecting the observations. The median and mean EuD are 5.13 and 5.15, respectively. Manhattan method calculates the distances between points with a different approach using only the absolute (positive) values of the points. In this dataset, the median and mean MD are 13.7 and 14.7, respectively.
K-means clustering and optimal number of clusters
set.seed(123)
k_max <- 10
twcss <- sapply(1:k_max, function(k){kmeans(boston_scaled, k)$tot.withinss})Visulaize the K-means clustering
library(ggplot2)
qplot(x = 1:k_max, y = twcss, geom = 'line')
Interpretation: I used the within cluster sum of squares (WCSS) to determine the optimal number of clusters. In the figure we are looking for an ‘elbow’ which demarks significant drop in rate of increase in WCSS. K=2 seems a good choice here.
Run K-means clustering with K=2 and its visulaization
km <-kmeans(boston_scaled, centers = 2)
pairs(boston_scaled, col = km$cluster)
Interpretation: Based on the figure, the variable ‘tax’in combination with most of the other variables distinguish the two clusters pretty well. Also, ’nox’&‘dis’ and ‘dis’&‘age’ distinguish the clusters better than the other variable combinations.
Bonus question
library(MASS)
data("Boston")
boston_scaled <- scale(Boston)
class(boston_scaled)## [1] "matrix"boston_scaled <- as.data.frame(boston_scaled)
km1 <- kmeans(boston_scaled, centers = 4)
pairs(boston_scaled, col = km1$cluster)
boston_scaled$k_cluster <- as.factor(km1$cluster)
lda.fit.1 <- lda(k_cluster ~ ., data = boston_scaled)
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "orange", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
classes <- as.numeric(boston_scaled$k_cluster)
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit.1, myscale = 1)
Interpretation: I performed K-means clustering on the standardized dataset with defind number of clusters=4. 202 observations belonged to cluster 1, 123 to C2, 128 to C3 and 53 to C4. Cluster 4 includes much lower proportion of the observations than the other 3 clusters and this could indicate that maybe 4 clusters is not the perfect solution. In the biplot, I can say that the linear combination of ‘indus & rad & black & zn’ variables are the most influential separators of the clusters.
Super-bonus question
model_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)
library(plotly)##
## Attaching package: 'plotly'## The following object is masked from 'package:ggplot2':
##
## last_plot## The following object is masked from 'package:MASS':
##
## select## The following object is masked from 'package:stats':
##
## filter## The following object is masked from 'package:graphics':
##
## layoutplot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = train$crime)